Problem 04 Evaluate $\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt$.
Solution 04 From Problem 01 | Division by t: $\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$
By first shifting property:
Problem 03 Find the value of $\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt$
Problem 02 Find the value of $\displaystyle \int_0^\infty \dfrac{\sin t ~dt}{t}$.
Problem 01 Evaluate $\displaystyle \int_0^\infty \dfrac{e^{-3t} - e^{-6t}}{t} ~ dt$
If $F(s) = \mathcal{L}\left\{ f(t) \right\}$, then $\displaystyle \int_0^\infty e^{-st} f(t) \, dt = F(s)$.
Taking the limit as $s \to 0$, then $\displaystyle \int_0^\infty f(t) \, dt = F(0)$ assuming the integral to be convergent.
Theorem If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, then
For first-order derivative: $\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$
For second-order derivative: $\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$
For third-order derivative: $\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$
For nth order derivative:
Division by $t$ If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, then,
provided $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$ exists.
Change of Scale Property If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, then,
Second Shifting Property If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, and $g(t) = \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$
then,
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